Problem: Solve for $x$ and $y$ using elimination. $\begin{align*}3x+3y &= -8 \\ -9x+y &= 4\end{align*}$
Answer: We can eliminate $y$ when its corresponding coefficients are negative inverses. Recalling our knowledge of least common multiples, multiply the top equation by $-1$ and the bottom equation by $3$ $\begin{align*}-3x-3y &= 8\\ -27x+3y &= 12\end{align*}$ Add the top and bottom equations. $-30x = 20$ Divide both sides by $-30$ and reduce as necessary. $x = -\dfrac{2}{3}$ Substitute $-\dfrac{2}{3}$ for $x$ in the top equation. $3( -\dfrac{2}{3})+3y = -8$ $-2+3y = -8$ $3y = -6$ $y = -2$ The solution is $\enspace x = -\dfrac{2}{3}, \enspace y = -2$.